It follows from, e.g., [1]} that also \(f^v\)  is submodular and that
\(P(f^v) &= \lbrace x \in P(f) \ | \ x \le v \rbrace ; &P^{\mathbb {Z}}(f^v) &= \lbrace x \in P^{\mathbb {Z}}(f) \ | \ x \le v \rbrace .\) 
